Basic question regarding Maxwells distribution

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The discussion centers on calculating the fraction of particles with speeds exceeding a threshold speed 'V0' using the Maxwell-Boltzmann distribution. To find this fraction (Nv/V), one must integrate the normalized distribution Nv from V0 to infinity. The cumulative distribution function provides a straightforward method, where the fraction of particles faster than V0 is determined by the formula 1 - D(V0). It is emphasized that the Maxwell distribution is rooted in classical mechanics, not quantum mechanics.

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In a system of particles described by the maxwells Boltzmann distribution . What would be the fraction (Nv/V) of the total number of particles 'N' having the speed above the given speed 'Vo'?
I have been puzzling through this answer for quite a while. I' am knew to Quantum mechanics so can anyone explain to me the solution to this problem?

Any help would be appreciated!
 
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If Nv is normalised, you need to integrate Nv dv with limits of V0 to infinity. Or, just read off the values from the cumulative distribution function.

Equation (3) here: http://mathworld.wolfram.com/MaxwellDistribution.html gives you a formula for the fraction of particles slower than x, so the fraction that are faster would be one minus this answer, ie. 1 - D(V0). Probably good to do this one on a computer.
 
Also, the Maxwell distribution is not quantummechanical in nature, it is a part of classical mechanics (more precisely, it's a part of classical equilibrium statistical mechanics).
 

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